Boundedness of $\dfrac{W_2(\mu_1+\varepsilon (\mu_2-\mu_1),\mu_1)}{\varepsilon}$ for 2 -Wasserstein metric

Question

Let $\mathcal{P}_2(\mathbb{R}^{n})$ the space of Borel probability measures of finite second moment in $\mathbb{R}^{n}$ equipped with the $2$-Wasserstein metric $W_2$. Let $\mu_1$, $\mu_2 \in \mathcal{P}_2(\mathbb{R}^{n})$ and $\varepsilon >0$, can we show $$\dfrac{W_2(\mu_1+\varepsilon (\mu_2-\mu_1),\mu_1)}{\varepsilon}<C(\mu_1,\mu_2)?$$ where $C(\mu_1,\mu_2)$ is some constant depending on $\mu_1$, $\mu_2$. I think it is true for $1$-Wasserstein metric by the Kantorovich duality.

Answer 1

First, consider the example where $\mu_1=\delta_0$ and $\mu_2=\delta_1$, then $W_2((1-\epsilon)\mu_1+\epsilon\mu_2,\mu_1)=\sqrt\epsilon$. Hence $\epsilon^{-1}W_2((1-\epsilon)\mu_1+\epsilon\mu_2,\mu_1)=(\sqrt\epsilon)^{-1}\to\infty$, and therefore there exists no upper bound not depending on $\epsilon$. However, if we consider $W_2^2$ things get different:

Here is an elementary approach which even works for general $W_p^p$:

Let $\pi_1$ be the optimal plan from $\mu_2$ to $\mu_1$ such that $$W_2^2(\mu_2,\mu_1)=\int|x-y|^2 \pi_1(dx,dy)$$ and let $\pi_0=(\text{id},\text{id})_{\text{#}}\mu_1$ the optimal transport plan from $\mu_1$ to itself.

Now an admissable transport plan $\pi_\epsilon$ from $(1-\epsilon)\mu_1+\epsilon\mu_2$ can be constructed by considering the mixture of $\pi_1$ and $\pi_0$, meaning $$\pi_\epsilon:=\epsilon \pi_1+(1-\epsilon)\pi_0.$$ One can easily verify that $\pi_\epsilon$ is a transport plan from $(1-\epsilon)\mu_1+\epsilon\mu_2$ to $\mu_1$: $$\begin{align*}\pi_\epsilon(\cdot\times\mathbb R^n)&=\epsilon\pi_1(\cdot\times\mathbb R^n)+(1-\epsilon)\pi_0(\cdot\times\mathbb R^n)=\epsilon\mu_2(\cdot)+(1-\epsilon)\mu_1(\cdot)\\ \pi_\epsilon(\mathbb R^n\times\cdot)&=\epsilon\pi_1(\mathbb R^n\times\cdot)+(1-\epsilon)\pi_0(\mathbb R^n\times\cdot)=\epsilon\mu_1(\cdot)+(1-\epsilon)\mu_1(\cdot)=\mu_1(\cdot)\end{align*}$$ Since $W_2^2$ is the infimum over all admissable transport plans, we get by definition $$\begin{align*}W_2^2((1-\epsilon)\mu_1+\epsilon\mu_2,\mu_1)&\leq \int |x-y|^2\pi_\epsilon(dx,dy)\\ &=(1-\epsilon)\int|x-y|^2\pi_0(dx,dy)+\epsilon\int|x-y|^2\pi_1(dx,dy) \\ &= \epsilon \int |x-y|^2\pi_1(dx,dy)=\epsilon W_2^2(\mu_2,\mu_1)\end{align*}$$ Consequently $$\epsilon^{-1}W_2^2((1-\epsilon)\mu_1+\epsilon \mu_2,\mu_1)\leq W_2^2(\mu_1,\mu_2)$$ Unfortunately, this naive estimate is not able to show that the limit as $\epsilon\searrow 0$ actually exists.

EDIT: After some discussion with Nate River and another acquaintance of our's, we have found a result which is relevant to your question: In short, if your $\mu_1,\mu_2$ are concentrated on a compact set $\Omega\subset \mathbb R^n$ and the Kantorovich potential for the transport from $\mu_1$ to itself is unique (up to addition of a constant), then the limit as $\epsilon \to 0$ (i.e. the derivative of the Wasserstein metric) turns out to be equal to 0. See for example Proposition 7.17 in Santambrogio's "Optimal Transport for Applied Mathematicians":

But remarkably, as the example in the beginning of my post shows ($\mu_1=\delta_0, \mu_2=\delta_1$), this is not always the case, since here clearly $\epsilon^{-1}W_2^2((1-\epsilon)\mu_1+\epsilon\mu_2,\mu_1)=1$. Note that in this case the Kantorovich potentials are not unique, which is why the aforementioned Theorem is not applicable.